CHAPTER 2: POLYNOMIAL AND RATIONAL FUNCTIONS


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1 CHAPTER 2: POLYNOMIAL AND RATIONAL FUNCTIONS 2.01 SECTION 2.1: QUADRATIC FUNCTIONS (AND PARABOLAS) PART A: BASICS If a, b, and c are real numbers, then the graph of f x = ax2 + bx + c is a parabola, provided a 0. ( ) = y If a > 0, it opens upward. If a < 0, it opens downward. Examples The graph of y = x 2 4x + 5 (with a = 1 > 0 ) is on the left. The graph of y = x 2 + 4x 3 (with a = 1 < 0 ) is on the right.
2 PART B: FINDING THE VERTEX AND THE AXIS OF SYMMETRY (METHOD 1) The vertex of the parabola [with equation] y = ax 2 + bx + c is ( h, k), where: xcoordinate = h = b 2a, and ycoordinate = k = f h ( ). The axis of symmetry, which is the vertical line containing the vertex, has equation x = h. Example (Does the formula for h look familiar? We will discuss this later.) Find the vertex of the parabola y = x 2 6x + 5. What is its axis of symmetry? Solution The vertex is ( h, k), where: h = b 2a = ( ) ( ) 2 6( 3) + 5 k = f 3 = 3 = 4 The vertex is ( 3, 4). ( ) = 3, and The axis of symmetry has equation x = 3. Since a = 1 > 0, we know the parabola opens upward. Together with the vertex, we can do a basic sketch of the parabola. 2.02
3 2.03 PART C: FINDING MORE POINTS Same Example: y = x 2 6x + 5, or f ( x) = x 2 6x + 5 Find the yintercept. Plug in 0 for x. Solve for y. In other words, find f ( 0). The yintercept is 5. (If the parabola is given by y = ax 2 + bx + c, then c, the constant term, is the yintercept. Remember that b was the yintercept for the line given by y = mx + b.) Find the xintercept(s), if any. Plug in 0 for y. Solve for x (only take real solutions). In other words, find the real zeros of f x ( ) = x 2 6x = x 2 6x + 5 ( )( x 1) 0 = x 5 The xintercepts are 1 and 5. x = 5 or x = 1 Note: Observe that f ( x) = x has no real zeros and, therefore, has no x intercepts on its graph. You can find other points using the PointPlotting Method (Notes 1.27). Symmetry helps!
4 2.04 PART D: PARABOLAS AND SYMMETRY Same Example Sketch the graph of y = x 2 6x + 5, or f ( x) = x 2 6x + 5. Clearly indicate: 1) The vertex 2) Which way the parabola opens 3) Any intercepts Make sure your parabola is symmetric about the axis of symmetry. Solution Point A is the vertex. Points B and C are the xintercepts. Point D is the yintercept. We get the bonus Point E by exploiting symmetry.
5 2.05 Observe that the xintercepts are symmetric about the axis of symmetry. This makes sense, because the zeros of f are given by the QF: x = b ± b2 4ac 2a The average of these zeros is b, which is the xcoordinate for the vertex and the 2a axis of symmetry. Technical Note: If the zeros of a quadratic f ( x) are imaginary, then they must have the same real part, which will be b 2a will be no xintercepts, however.. The axis of symmetry will still be x = b 2a. There Technical Note: The xintercept on the right (if any) does not always correspond to the + case in the QF. Remember, a can be negative.
6 2.06 PART E: STANDARD FORM FOR THE EQUATION OF A PARABOLA; FINDING THE VERTEX AND THE AXIS OF SYMMETRY (METHOD 2) The graph of y = a x h y = ax 2, but its vertex is ( h, k). Example ( ) 2 + k is a parabola that has the same shape as the parabola Remember translations from Section 1.6. The parabola y = ax 2, which has vertex 0, 0 ( ), is shifted h units horizontally and k units vertically to obtain the new parabola. If y = a( x h) 2 + k is written out in the form y = ax 2 + bx + c, then the a s are, in fact, the same. Rewrite y = 3x x + 11 in the form y = a( x h) 2 + k, and find the vertex. Solution Step 1: Group the quadratic and linear terms and factor out a from the group. ( ) + 11 ( ) + 11 y = 3x x y = 3 x 2 + 4x Step 2: Complete the Square (CTS) and Compensate. Take the coefficient of x inside the (), halve it, and square the result. This is the number we add inside the (). Half of 4 is 2, and the square of 2 is 4. ( ) ( ) y = 3 x 2 + 4x PST
7 2.07 Step 3: Factor our Perfect Square Trinomial (PST) as the square of a binomial, and simplify. y = 3( x + 2) 2 1 The vertex is ( 2, 1). Warning: Don t forget the first "." Remember that the xcoordinate of the vertex must make ( x + 2) equal to 0. This makes sense, because the vertex is an extreme point, and 0 is as extreme a square can get. The axis of symmetry is x = 2. a = 3 > 0, so the parabola opens upward. Note: A key advantage that the form y = a( x h) 2 + k has over the form y = ax 2 + bx + c is that the vertex is easier to find using the first form. Also, xintercepts (but not the yintercept) may be easier to find.
8 2.08 PART F: GIVEN SOME POINTS, FIND THE PARABOLA PASSING THROUGH THEM Two distinct points determine a line. Three noncollinear points (i.e., points that do not lie on a straight line) determine a parabola. If we know the vertex and one other point on the parabola, then we can get a third point automatically by exploiting symmetry. Technical Note: If you successively plug in the coordinates of the three points into the form y = ax 2 + bx + c, you will obtain a system of three linear equations in three unknowns (a, b, and c). Have fun! If we know the vertex, then the form y = a( x h) 2 + k makes matters easier for us. Example Find an equation for the parabola that has vertex 4, 2 ( 1, 29). Solution Using the vertex: ( ) 2 + k ( ) y = a x h y = a x 4 Plug in ( x = 1, y = 29) and solve for a: 29 = a( 1 4) = 9a = 9a a = 3 ( ) and that contains the point
9 2.09 An equation for the parabola is: y = 3( x 4) Here s a graph: Note: In applications, you can think of this as modeling a quadratic function. We will see parabolas again in Section 10.2.
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